Frequent values

Frequent values

Description
You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers ai , ... , aj.
Input
The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a1 , ... , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the query. 


The last test case is followed by a line containing a single 0.
Output
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.
Sample Input
10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0
Sample Output
1
4
3
Hint

A naive algorithm may not run in time!

首先给定一个非降序的数列，然后进行查询操作，问范围[a,b]内出现频率最高的数字出现了多少次。

因为通过巧妙的转换，能使它成为一个RMQ问题。

num[i]是原数列，num2[i]是转换后的数列,num2[i]代表与num[i]相同的左边的节点数（包括自己），例如，

（-1 -1 1 1 1 1 3 10 10 10）==>（1 2 1 2 3 4 1 1 2 3）。

这时候因为一个数字出现的频率已经直接出现在num2[i]数列中，我们就可以对num2[i]使用RMQ算法了。不过因为相同的连续的数可能被切成了两段，所以在边界上需要特殊处理。我们把查询范围[a,b]内的数分为二段，第一段为：num[a] ~num[i],其中num[a] > 1，num[i] != 1,num[a] <num[j] <...<num[i]。第二段为num[i+1]...num[b],此段可以用RMQ.。

代码如下：

#include "stdio.h"#include "math.h"#define max(a,b) (a > b ? a : b)#define MAXN 100000int num[MAXN];int num2[MAXN];int dp[MAXN][20];void create (int n) {int i,j,t;for (i = 1;i <= n;i++) {dp[i][0] = num2[i];}t = log(n) / log(2.0);for (j = 1;j <= t;j++) {for (i = 1;i + (1 << j) - 1 <= n;i++) {dp[i][j] = max(dp[i][j-1],dp[i + (1 << (j-1))][j-1]);}}}int getMax (int a,int b) {int k = (int)(log(b - a + 1.0) / log(2.0));return max (dp[a][k],dp[b-(1<< k) + 1][k]);}int query (int a,int b) {if (num[a] == num[b]) {return b - a + 1;}else {int left,_a;int i = a;while (num2[i] != 1) {i++;}_a = i;if (i == a) {left = 0;}else {left = num2[i-1] - num2[a] + 1;}return max(left,getMax(_a,b));}}void print (int n,const int *array) {int i;for (i = 1;i <= n;i++) {printf ("%4d ",array[i]);}printf ("\n");}int main () {int n,q,i,j,a,b;int t;while (scanf ("%d",&n), n != 0) {scanf ("%d",&q);num[0] = 1;num2[0] = 0;for (i = 1;i <= n;i++) {scanf ("%d",&num[i]);}//print (n,num);for (i = 1;i <= n;i++) {if (num[i] == num[i-1]) {num2[i] = num2[i-1] + 1;}else {num2[i] = 1;}}//print (n,num2);create(n);for (i = 1;i <= q;i++) {scanf ("%d%d",&a,&b);t = query (a,b);printf("%d\n",t);}}return 0;}